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From Exercise 11 data. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron ? Explain. |
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Answer» Solution :Given `B=6.5 G=6.5 xx10^(-4)T, V=4.8xx10^(6) m//s, e=1.6xx10^(-19)C` `m_(e)=9.1xx10^(-3) KG` `(nV^(2))/(r )= Q V B rArr (mV)/(r )=q B` If angular VELOCITY of electron is `omega`, then `V=r omega` `(m r(omega))/(r )=qB` `omega =(qB)/(m)` `2pi n=(qB)/(m) rArr n=(qB)/(2 pi m)` Frequency of revolution of electron in orbit `v=(Bq)/(2pi m)=(Be)/(2pi m_(e))=(6.5xx10^(-4)xx1.6xx10^(-19))/(2xx3.14xx9.1xx10^(-31))=18.18xx 10^(6) Hz.` |
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